Câu 1:
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
Câu 2:
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PTHH ta có tỉ lệ:
\(\dfrac{0,1}{4}=0,025< \dfrac{0,2}{5}=0,04\)
=> P hết. \(O_2\) dư. => tính theo \(n_P\)
Theo PT ta có: \(n_{P_2O_5}=\dfrac{0,1.2}{4}=0,05\left(mol\right)\)
=> \(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)